Question

18% of all Americans live in poverty. If 44 Americans are randomly selected, find the probability that

a. Exactly 9 of them major in STEM.  
b. At most 12 of them major in STEM.  
c. At least 8 of them major in STEM.  
d. Between 9 and 13 (including 9 and 13) of them major in STEM.

Answer 1

Solution:

Given that,

P = 0.18

1 - P = 0.82

n = 44

Here, BIN ( n , P ) that is , BIN (44 , 0.18)

then,

n*p = 7.92 > 5

n(1- P) = 36.08 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 7.92

Standard deviation = =n*p*(1-p) = 6.4944

We using continuity correction factor

a)

P(X = a) = P( a - 0.5 < X < a + 0.5)

P(8.5 < x < 9.5) = P((8.5 - 7.92)/ 6.4944 ) < (x - ) /  < (9.5 - 7.92) / 6.4944 ) )

= P(0.23 < z < 0.62)

= P(z < 0.62) - P(z < 0.23)

= 0.7324 - 0.5910

Probability = 0.1414

b)

P( X a ) = P(X < a + 0.5)

P(x < 12.5) = P((x - ) / < (12.5 - 7.92) /6.4944 )

= P(z < 1.80)

Probability = 0.9641

c)

P(X a ) = P(X > a - 0.5)

P(x > 7.5) = 1 - P(x < 7.5)

= 1 - P((x - ) / < (7.5 - 7.92) / 6.4944)

= 1 - P(z < -0.16)

= 1 - 0.4364

= 0.5636

Probability = 0.5636

d)

P(9 X 13) = P( a - 0.5 < X < b + 0.5)

P(8.5 < x < 13.5) = P((8.5 - 7.92)/ 6.4944) < (x - ) /  < (13.5 - 7.92) / 6.4944) )

= P(0.23 < z < 2.19)

= P(z < 2.19) - P(z < 0.23)

= 0.9857 - 0.5910

Probability = 0.3947